Math  /  Algebra

Question3. Use CRT to solve the following system of linear congruences: 2x=5(mod7)4x=2(mod6)x=3(mod5)\begin{aligned} 2 x & =5(\bmod 7) \\ 4 x & =2(\bmod 6) \\ x & =3(\bmod 5) \end{aligned} (Hint: First determine how many unique solutions modulo 210 exist. Why?)

Studdy Solution

STEP 1

1. We are given a system of linear congruences to solve using the Chinese Remainder Theorem (CRT).
2. The moduli are 77, 66, and 55, which are pairwise coprime.
3. We need to find a solution xx that satisfies all three congruences simultaneously.
4. The hint suggests determining the number of unique solutions modulo 210210, which is the product of the moduli.

STEP 2

1. Determine the product of the moduli and why it is important.
2. Solve each congruence individually.
3. Use the Chinese Remainder Theorem to find a common solution.
4. Verify the solution and determine the number of unique solutions modulo 210.

STEP 3

Calculate the product of the moduli:
N=7×6×5=210 N = 7 \times 6 \times 5 = 210
Since the moduli are pairwise coprime, the Chinese Remainder Theorem guarantees a unique solution modulo 210210.

STEP 4

Solve the first congruence:
2x5(mod7) 2x \equiv 5 \pmod{7}
Find the multiplicative inverse of 22 modulo 77. The inverse is a number bb such that 2b1(mod7)2b \equiv 1 \pmod{7}.
Test values to find bb:
2×481(mod7) 2 \times 4 \equiv 8 \equiv 1 \pmod{7}
Thus, the inverse is 44.
Multiply both sides of the congruence by 44:
x206(mod7) x \equiv 20 \equiv 6 \pmod{7}

STEP 5

Solve the second congruence:
4x2(mod6) 4x \equiv 2 \pmod{6}
Simplify by dividing by the greatest common divisor, which is 22:
2x1(mod3) 2x \equiv 1 \pmod{3}
Find the multiplicative inverse of 22 modulo 33. The inverse is 22 because:
2×241(mod3) 2 \times 2 \equiv 4 \equiv 1 \pmod{3}
Multiply both sides by 22:
x2(mod3) x \equiv 2 \pmod{3}

STEP 6

Solve the third congruence:
x3(mod5) x \equiv 3 \pmod{5}
This congruence is already simplified.

STEP 7

Use the CRT to find a common solution. We have:
x6(mod7) x \equiv 6 \pmod{7} x2(mod3) x \equiv 2 \pmod{3} x3(mod5) x \equiv 3 \pmod{5}
Calculate the solution using the method of successive substitutions or direct application of CRT formulas.

STEP 8

Calculate the individual terms:
- For x6(mod7)x \equiv 6 \pmod{7}, N1=2107=30N_1 = \frac{210}{7} = 30, find M1M_1 such that 30M11(mod7)30M_1 \equiv 1 \pmod{7}. M1=1M_1 = 1.
- For x2(mod3)x \equiv 2 \pmod{3}, N2=2106=35N_2 = \frac{210}{6} = 35, find M2M_2 such that 35M21(mod3)35M_2 \equiv 1 \pmod{3}. M2=2M_2 = 2.
- For x3(mod5)x \equiv 3 \pmod{5}, N3=2105=42N_3 = \frac{210}{5} = 42, find M3M_3 such that 42M31(mod5)42M_3 \equiv 1 \pmod{5}. M3=3M_3 = 3.
Combine these using CRT:
x6×30×1+2×35×2+3×42×3(mod210) x \equiv 6 \times 30 \times 1 + 2 \times 35 \times 2 + 3 \times 42 \times 3 \pmod{210}
Calculate each term:
6×30×1=180 6 \times 30 \times 1 = 180 2×35×2=140 2 \times 35 \times 2 = 140 3×42×3=378 3 \times 42 \times 3 = 378
Sum them up:
x180+140+37869868(mod210) x \equiv 180 + 140 + 378 \equiv 698 \equiv 68 \pmod{210}

STEP 9

Verify the solution:
- Check x6(mod7)x \equiv 6 \pmod{7}: 686(mod7)68 \equiv 6 \pmod{7} - Check x2(mod3)x \equiv 2 \pmod{3}: 682(mod3)68 \equiv 2 \pmod{3} - Check x3(mod5)x \equiv 3 \pmod{5}: 683(mod5)68 \equiv 3 \pmod{5}
All checks are correct. Therefore, the solution is x68(mod210)x \equiv 68 \pmod{210}.

Was this helpful?

Studdy solves anything!

banner

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

ParentsInfluencer programContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord